In SO₂$\text{SO}_2$, NO₂^-$\text{NO}_2^-$ and N₃^-$\text{N}_3^-$ the hybridizations at the central atom are respectively:

Related Formula

Core Logic

Let's perform steric calculations for each species:

• SO₂$\text{SO}_2$: Central sulfur atom has 6 valence electrons, forms 2 σ$2\,\sigma$-bonds (and 2 π$2\,\pi$-bonds) with oxygen, leaving 1 lone pair. Steric number = 2 + 1 = 3 sp²$= 2 + 1 = 3 \implies sp^2$.
• NO₂^-$\text{NO}_2^-$: Central nitrogen atom has 5 valence electrons + 1$+ 1$ from negative charge = 6$= 6$. It forms 2 σ$2\,\sigma$-bonds, leaving 1 lone pair. Steric number = 2 + 1 = 3 sp²$= 2 + 1 = 3 \implies sp^2$.
• N₃^-$\text{N}_3^-$ (Azide ion): Linear configuration structure can be drawn as:

The central nitrogen has 2 σ$2\,\sigma$-bonds and 0 lone pairs. Steric number = 2 + 0 = 2 sp$= 2 + 0 = 2 \implies sp$.

Step 1: Geometry Outlines

The individual orbital fields are represented visually:

Hence, hybridizations follow the order: sp²$sp^2$, sp²$sp^2$, and sp$sp$.

Pattern Recognition

Steric short tracking: Species with linear structures like CO₂, N₂O, N₃^-$\text{CO}_2, \text{N}_2O, \text{N}_3^-$ possess central atoms that are always sp$sp$ hybridized due to the requirement of two opposing σ$\sigma$-bonds.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure